Examine the differentiability of f(x) = [x] at x =1
Answers
Answer:
We have,
f(x)=
⎩
⎪
⎪
⎨
⎪
⎪
⎧
2x+3,
x+1,
x+2,
if
if
if
−3≤x<−2
−2≤x<0
0≤x≤1
.
Since, the polynomial function are continuous and differentiable at everywhere. So, f(x) is differentiable on x∈[−3,−2),x∈(−2,0) and x∈(0,1].
We need to check the differentiablility at x=−2 and x=0.
Differentiablity at x=−2,
(LHD at x=−2)=lim
x→ −2
−
x−(−2)
f(x)−f(−2)
=lim
x→ −2
−
x+2
2x+3−(−1)
=lim
x→ −2
−
x+2
2x+4
=lim
x→ −2
−
x+2
2(x+2)
=2
Now,
(RHD at x=−2)=lim
x→ −2
+
x−(−2)
f(x)−f(−2)
=lim
x→ −2
+
x+2
x+1−(−1)
=lim
x→ −2
+
x+2
x+2
=1
Therefore,
(LHD at x=−2)
=(RHD at x=−2)
So, f(x) is not differentiable at x=−2.
Differentiablity at x=0,
(LHD at x=0)=lim
x→ 0
−
x−0
f(x)−f(0)
=lim
x→ 0
−
x
x+1−2
=lim
x→ 0
−
x
x−1
→∞
Now,
(RHD at x=0)=lim
x→ 0
+
x−0
f(x)−f(0)
=lim
x→ 0
+
x
x+2−2
=lim
x→ 0
+
x
x
=1
Therefore,
(LHD at x=0)
=(RHD at x=0)
So, f(x) is not differentiable at x=0.